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By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt.

By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt. . Reinforced Concrete Design-3 Flexural Design of Beams. Working Stress Design Method . Design of beam by Working Stress method .

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By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt.

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  1. ByDr. Attaullah ShahSwedish College of Engineering and Technology Wah Cantt. Reinforced Concrete Design-3 Flexural Design of Beams

  2. Working Stress Design Method

  3. Design of beam by Working Stress method Design a simply supported beam of span 20 feet carrying a live load of 1.5 K/ft and dead load of 2 K/ft. the material stresses are fc=1300 psi and fs=22300 psi. Es =29x106 psi and Ec= 36x105 psi W=D+L = 1.5+2 = 3.5K/ft Mmax= wl2/8 = 3.5*20*20/8= 175 ft-K=175*12=2100 in-K The sizes of the beams: The modular ratio n = Es/Ec= 9x106 /36x105= 8 = = 0.318 j= 1-k/3 = 1-0.318/3 = 0.894 ; bd2 = M/0.5fc k j = 2100*1000/(0.5*1300*0.318*0.894=11364 in 3 Using d/b = 1.5 we get d=1.5b or b(1.5b) 2 = 11364 or 2.25b 3 = 11364 or b = 17.1 in. We may take b = 18 in Now 18d 2 = 11364 or d 2 = 11364/18 or d= 25 say 12 in Over all depth = 25+1.5 in ( cover)+0.5= 127 in ( beam size 18 inX27in) Design of steel As = M/fsjd = 2100*1000/22300*0.894*25 = 4.21 in2 Using # 8 bars No of bars = 4.21/0.87 = 5 bars

  4. Safety Margins

  5. ACI Provisions: • Design strength > Required strength • Su = 1.2 DL + 1.6 LL

  6. For fc’< 4000 psi ß= 0.85 For fc’ > 4000 psi ß = 0.85 - 0.05 (fc’-4000)/1000 0.65> ß>0.85

  7. Under reinforced Reinforced Concrete beams • The compression failure occurs due to crushing of concrete, which is sudden and give no warning. • The tension failure is caused by yielding of steel bars. • Due to yielding of concrete beams having large deflections, the cracks can be observed and preventive measures can be taken. • The steel ratio is taken less than balanced steel to ensure ductile failure • 200/fy

  8. Strength Reduction Factor ϕ

  9. Flexural Design of Rectangular beam Design a rectangular simply supported beam subjected to computed dead load of 1.27 k/ft and live load of 2.15 k/ft. fc’ = 4000 psi and fy=60,000 psi.

  10. Example: 3.7 No further iteration is required. Use As=1.49in2 2#8 bars (No.25) bars will be used Check whether steel ratio is less than maximum allowed by the ACI Code.

  11. Example 3.8

  12. Use of Design Aids • We can directly use design the beam by using design charts. • The optimum reinforcement is selected. • Set the required strength Mu equal to the design strength ϕMn • Mn = ϕRbd2 • Using Design Table A.4 select the appropriate reinforcement ratio between ρmin and ρmax Often ρ = 0.60 ρmax is more economical. • From Table A.5 for specified material strengths and selected reinforcement, find flexural resistance factor R then • bd2 = Mn /ϕR • Select the appropriate width and depth of beam. ( d=2-3b) • Calculate the areas of steel As= ρbd • Select the number and size of bars ( prefer larger sizes) • Alternatively we can assume the values of b and d and determine R =Mn/ ϕbd2 • Using Table A.5 find the reinforcement ratio ρ< ρmax that will provide the required value of R • As= ρbd

  13. Table A.5 a Flexural Resistance Factor

  14. Assignment 2. Group 1 Group 2: Group 3:

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